Trả lời:

a, \(\left(x^2-2y\right)\left(x^4+2x^2y+4y^2\right)-x^3\left(x-y\right)\left(x^2+xy+y^2\right)+8y^3\)

\(=\left(x^2\right)^3-\left(2y\right)^3-x^3\left(x^3-y^3\right)+8y^3\)

\(=x^6-8y^3-x^6+x^3y^3+8y^3\)

\(=x^3y^3\)

b, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3+7\)

\(=x^3-8-\left(x^3-3x^2+3x-1\right)+7\)

\(=x^3-8-x^3+3x^2-3x+1+7\)

\(=3x^2-3x\)

c, \(x\left(x+2\right)\left(2-x\right)+\left(x+3\right)\left(x^2-3x+9\right)\)

\(=x\left(4-x^2\right)+x^3+27\)

\(=4x-x^3+x^3+27\)

\(=4x+27\)

\(^{ }\)

\(\frac{a}{b}+\frac{b}{a}\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}=2\sqrt{1}=2\)

dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}\frac{a}{b}=\frac{b}{a}\\a+b=4\end{cases}}\)

vậy \(MIN=2\)

Ta có: a/b+b/a=\(\frac{a^2+b^2}{ba}\)= \(\frac{\left(a+b\right)^2}{ba}-2\)=16/ab-2

hay để a/b và b/a nhỏ nhất thì ba lớn nhất và khác 0 (rồi giờ bn tìm ba thôi, đừng bấm sai vì mình chưa ra kq nhé)

Giải

1/(2.4) + 1/(4.6) + … + 1/[(2x – 2).2x] = 1/8
=> 2/(2.4) + 2/(4.6) + ...+ 2/[(2x - 2).2x] = 2/8
 =>1-1/4+1/4-1/6+...+1/(2x-2) - 1/2x = 2/8
 =>1 - 1/2x = 2/8
 =>1/2x = 1 - 2/8
 =>1/2x = 6/8 = 3/4
=>1.4 = 2.x.3
=>4 = 6x
=> x thuộc rỗng
 Vậy x thuộc rỗng

Trả lời:

\(A=\left(2x+5\right)\left(4x^2-10x+25\right)-\left(16+8x^3\right)\)

\(=8x^3+125-16-8x^3\)

\(=109\)

chiu ba bao tui lam lo lon ak

Bài 3B : 

a, \(\left(x+4\right)^2-\left(2x+1\right)^2=3\left(x-3\right)\)

\(\Leftrightarrow\left(x+4-2x-1\right)\left(x+4+2x+1\right)=3\left(x-3\right)\)

\(\Leftrightarrow-\left(x-3\right)\left(3x+5\right)=3\left(x-3\right)\Leftrightarrow\left(x-3\right)\left(-5-3x\right)-3\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(-8-3x\right)=0\Leftrightarrow x=-\frac{8}{3};x=3\)

b, \(x^3-8=2x^2-4x\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-2x\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2+4>0\right)=0\Leftrightarrow x=2\)

c, \(x^3-6x^2+8x=0\Leftrightarrow x\left(x^2-6x+8\right)=0\)

\(\Leftrightarrow x\left(x^2-6x+9-1\right)=0\Leftrightarrow x\left[\left(x-3\right)^2-1\right]=0\)

\(\Leftrightarrow x\left(x-4\right)\left(x-2\right)=0\Leftrightarrow x=0;x=2;x=4\)

a, \(x^2-6x+9=\left(x-3\right)^2\)

b, \(x^2-12x+36=\left(x-4\right)^2\)

c, \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)

d, \(x^2-x+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2\)

e, \(x^4-8x^2+16=\left(x^2-4\right)^2=\left[\left(x-2\right)\left(x+2\right)\right]^2\)

f, \(x^4-81=\left(x^2-9\right)\left(x^2+9\right)=\left(x-3\right)\left(x+3\right)\left(x^2+9\right)\)

g, \(\left(4x+5\right)^2-\left(5x+4\right)^2=\left(4x+5-5x-4\right)\left(4x+5+5x+4\right)=9\left(1-x\right)\left(x+1\right)\)

h, \(\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(-x-2\right)^2\)

\(=\left(2x-3\right)^2-2\left(2x-3\right)\left(x+2\right)+\left(x+2\right)^2\)

\(=\left(2x-3-x-2\right)^2=\left(x-5\right)^2\)

a) x² - 6x + 9 = (x-3)²

b) x² - 12 + 36 = (x-6)²

c) 9x² - 25 = (3x - 25)(3x + 25)

d) x² - x + 1/4 = (x - 1/2)²

a, \(x^2-6x+9=4< =>\left(x-3\right)^2=4< =>\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(< =>\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b,\(x^2\left(x-3\right)-4\left(x-3\right)=0< =>\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(< =>\orbr{\begin{cases}x=2\\x=-2\end{cases}orx=3}\)

c nhường mấy bn khácccc

a) x^2-6x+9=4.

 x=1, x=5

b) x^2(x-3)-(4X-12)=0

x=-2, x=2, x=3

c) (2x+3)^2-4(x+2)^2=12

x=-19/4

(x+2)(x+3)(x+4)(x+5)=24

x=-6,

x=-1;

x = -(căn bậc hai(3)căn bậc hai(5)i+7)/2

;x = (căn bậc hai(3)căn bậc hai(5)i-7)/2;

nha bạn chúc bạn học tốt nha 

(x + 2)(x + 3)(x + 4)(x + 5) = 24

<=> [(x + 2)(x + 5][(x + 3)(x + 4] = 24

<=> (x² + 7x + 10)(x² + 7x + 12) - 24 = 0

<=> (x² + 7x + 11 - 1)(x² + 7x + 11 + 1) - 24 = 0

<=> (x² + 7x + 11)² - 25 = 0 

<=> (x² + 7x + 16)(x² + 7x + 6) = 0

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left[\left(x+\frac{7}{2}\right)^2+\frac{15}{4}\right]=0\)

<=> (x + 1)(x + 6) = 0 

<=> \(\orbr{\begin{cases}x+1=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-6\end{cases}}\)

Vậy \(x\in\left\{-1;-6\right\}\)